$$(x^6-1)^2(x^2+3)^3 = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = 0.5+0.86603i \\[1 em] x_4 = 0.5-0.86603i & x_5 = -0.5+0.86603i & x_6 = -0.5-0.86603i \\[1 em] x_7 = 1.73203i & x_8 = -1.73203i & x_9 = -2.0E-5+1.73206i \\[1 em] x_10 = -2.0E-5-1.73206i & x_11 = 2.0E-5+1.73206i & x_12 = 2.0E-5-1.73206i \end{matrix} $$
Explanation
$$ \begin{aligned} (x^6-1)^2(x^2+3)^3 &= 0&& \text{simplify left side} \\[1 em](x^{12}-2x^6+1)(x^6+9x^4+27x^2+27) &= 0&& \\[1 em]x^{18}+9x^{16}+27x^{14}+25x^{12}-18x^{10}-54x^8-53x^6+9x^4+27x^2+27 &= 0&& \\[1 em] \end{aligned} $$
This page was created using
Polynomial Equations Solver